Question
Prove that $A(4, 3), B(6, 4), C(5, 6)$ and $D(3, 5)$ are the angular points of a square.
✓
Answer
Now
$AB = \sqrt{(4-6)^2+(3-4)^2}$
$=\sqrt{4+1}=\sqrt{5} \text { units }$
$B C=\sqrt{(6-5)^2+(4-6)^2}=\sqrt{1+4}$
$B C=\sqrt{5} \text { units. }$
$C D=\sqrt{(5-3)^2+(6-5)^2}=\sqrt{4+1}$
$C D=\sqrt{5} \text { units }$
$\text { Also DA }=\sqrt{(4-3)^2+(3+5)^2}$
$=\sqrt{1+4}=\sqrt{5}$
$\text { DA }=\sqrt{5} \text { units }$
$\text { So } A B=B C=C D=D A .$
Now slope of $A B=m_1-\frac{4-3}{6-4}=\frac{1}{2}$
Slope of $B C=m_2=\frac{6-4}{5-6}=\frac{2}{-1}$
Slope of $C A=m_3=\frac{5-6}{3-5}=\frac{1}{2}$
Slope of DA $=m_4=\frac{5-3}{3-4}=\frac{2}{-1}$
Since $m_1 = m_3$ and $= m_2 = m_4$
So $AB \| CD$
and $BC \| DA.$
Therefore, $AB \perp BC$
$\therefore ABCD$ is a square.
Hence proved.
$AB = \sqrt{(4-6)^2+(3-4)^2}$
$=\sqrt{4+1}=\sqrt{5} \text { units }$
$B C=\sqrt{(6-5)^2+(4-6)^2}=\sqrt{1+4}$
$B C=\sqrt{5} \text { units. }$
$C D=\sqrt{(5-3)^2+(6-5)^2}=\sqrt{4+1}$
$C D=\sqrt{5} \text { units }$
$\text { Also DA }=\sqrt{(4-3)^2+(3+5)^2}$
$=\sqrt{1+4}=\sqrt{5}$
$\text { DA }=\sqrt{5} \text { units }$
$\text { So } A B=B C=C D=D A .$
Now slope of $A B=m_1-\frac{4-3}{6-4}=\frac{1}{2}$
Slope of $B C=m_2=\frac{6-4}{5-6}=\frac{2}{-1}$
Slope of $C A=m_3=\frac{5-6}{3-5}=\frac{1}{2}$
Slope of DA $=m_4=\frac{5-3}{3-4}=\frac{2}{-1}$
Since $m_1 = m_3$ and $= m_2 = m_4$
So $AB \| CD$
and $BC \| DA.$
Therefore, $AB \perp BC$
$\therefore ABCD$ is a square.
Hence proved.
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